数位dp

ll dfs(int*num, int n, int pre, int pos, bool limit, bool lead){//try to fill the pos bit
    if(pos==n)return 1;
    else{
        if(!limit&&dp[pre==6][num[pos]][n-pos][lead]!=-1)
            return dp[pre==6][num[pos]][n-pos][lead];
        else {
            int upper=limit ? num[pos] : 9;
            ll ans=0;
            for(int i=0;i<=upper;i++){
                if(i==4)continue;
                if(pre==6&&i==2)continue;
                ans+=dfs(num,n,i,pos+1,limit&&i==upper,lead||i!=0);
            }
            if(!limit) dp[pre==6][num[pos]][n-pos][lead]=ans;
            return ans;
        }
    }
}